Let $z_i \in \mathbb{R}$, for every $i \in \mathbb{Z}_{[1,n]}$ be a set of scalars. Let $a_i \in \mathbb{R}$, for every $i \in \mathbb{Z}_{[1,n]}$ be a set of scalars such that $\sum_{i=1}^na_i=1$ and $a_i\geq 0$. How can I show, possibly in the most intuitive way, that $$\sum_{i=1}^na_iz_i \leq \max_{i\in \mathbb{Z}_{[1,n]}}z_i\,.$$
Specifically, I don't want to use the fact that it is a convex combination, so that the result would hold by definition.
Hint: the $i$th term in your sum is bounded above by $a_i\max z_i$, because each $a_i$ is non-negative. (Without this non-negativity requirement, the claim doesn't hold: consider $z_1=-1$, $z_2=0$, $a_1=-1$, $a_2=2$.)